Exercise 4.1.11 (Adjoined Expansion)

Proof. Let $F$ be a subfield of $M$ such that $F\supset K\cup (Y\cup Z)$. Then $K\subset F$, and $Y\cup Z\subset F$, thus $Y\subset F$ and $Z\subset F$. Moreover $K\subset F$, and $Y\subset F$, thus $K\cup Y\subset F$. This implies that $K(Y)\subset F$, because $K(Y)$ is the smallest subfield of $M$ which contains $K\cup Y$. So $K(Y)\cup Z\subset F$.

Conversely, if $(K(Y))(Z)\subset F$ then $K(Y)\subset F$ and $Z\subset F$, so $F$ contains $K,Y$ and $Z$, that is $F\supset K\cup (Y\cup Z)$. We have proved, for all subfields $F$ of $M$, that

Since $K(Y\cup Z)$ is the least subfield of $M$ such that $F\supset K\cup (Y\cup Z)$, and $(K(Y))(Z)$ the least subfield of $M$ such that $F\supset K(Y)\cup (Z)$, this two subfields are equals:

If we write $ℱ$ the set of subfields of $M$,

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